hdu 5586 Sum 基础dp

Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

There is a number sequence A1,A2....An,you can select a interval [l,r] or not,all the numbers Ai(l≤i≤r) will become f(Ai).f(x)=(1890x+143)mod10007.After that,the sum of n numbers should be as much as possible.What is the maximum sum?

 

Input

There are multiple test cases.
First line of each case contains a single integer n.(1≤n≤105)
Next line contains n integers A1,A2....An.(0≤Ai≤104)
It's guaranteed that ∑n≤106.

 

Output

For each test case,output the answer in a line.

 

Sample Input

2 10000 9999 5 1 9999 1 9999 1

 

Sample Output

19999 22033

 

Source

BestCoder Round #64 (div.2)

题目链接：点击传送

思路：基础dp；

　　　dp[i][0]表示dp前边没有改成f的总和

　　　dp[i][1]表示可以继续改成f的总和；

　　　dp[i][2]表示不可以继续改成f的总和；

dp[i][0]=dp[i-1][0]+a[i];
dp[i][1]=max(dp[i-1][1]+b[i],dp[i-1][0]+b[i]);
dp[i][2]=max(dp[i-1][2]+a[i],dp[i-1][1]+a[i]);

　　　b=f(a);

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=1e5+30010,M=1e6+10,inf=2147483647;
const ll INF=1e18+10,mod=2147493647;
ll a[N],b[N],dp[N][4];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]),b[i]=(1890*a[i]+143)%10007;
        dp[0][0]=dp[0][1]=dp[0][2]=0;
        for(int i=1;i<=n;i++)
        {
            dp[i][0]=dp[i-1][0]+a[i];
            dp[i][1]=max(dp[i-1][1]+b[i],dp[i-1][0]+b[i]);
            dp[i][2]=max(dp[i-1][2]+a[i],dp[i-1][1]+a[i]);
        }
        printf("%lld
",max(dp[n][2],max(dp[n][0],dp[n][1])));
    }
    return 0;
}